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Professional Application Ernest Rutherford (the first New

Physics: Principles with Applications | 6th Edition | ISBN: 9780130606204 | Authors: Douglas C. Giancoli ISBN: 9780130606204 3

Solution for problem 49PE Chapter 8

Physics: Principles with Applications | 6th Edition

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Physics: Principles with Applications | 6th Edition | ISBN: 9780130606204 | Authors: Douglas C. Giancoli

Physics: Principles with Applications | 6th Edition

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Problem 49PE

Professional Application Ernest Rutherford (the first New Zealander to be awarded the Nobel Prize in Chemistry) demonstrated that nuclei were very small and dense by scattering helium-4 nuclei (4He) from gold-197 nuclei (197 Au). The energy of the incoming helium nucleus was 8.00×10?13 J , and the masses of the helium and gold nuclei were 6.68×10?27 kg and 3.29×10?25 kg , respectively (note that their mass ratio is 4 to 197). (a) If a helium nucleus scatters to an angle of 120º during an elastic collision with a gold nucleus, calculate the helium nucleus’s final speed and the final velocity (magnitude and direction) of the gold nucleus. (b) What is the final kinetic energy of the helium nucleus?

Step-by-Step Solution:

Step-by-step solution In this problem we have to find the final speed and the final velocity of the gold nucleus and expression to find the final kinetic energy of helium Step 1 of 9 (a) The expression to find the kinetic energy is, Rearrange the expression for velocity. Here, is kinetic energy, is mass and is initial velocity of Helium. Step 2 of 9 Substitute, for and for in the expression of velocity, 7 Hence the initial velocity is 1.548 × 10 m/s Step 3 of 9 As collision is elastic; apply conservation of momentum along x -axis, Here, are the masses of helium and gold respectively, is initial velocity of helium, are scattered angle for helium and gold respectively, is the final velocity of gold and is the final velocity of gold. he expression for conservation of momentum along y-axis is: Rearranging the above expression of conservation of momentum respectively, Step 4 of 9 Take square of both of them respectively and add the expressions, Now, from the trigonometric relation: Use trigonometric relation in the above expression; the new expression is, Step 5 of 9 Now, the conservation of internal kinetic energy gives: Rearrange the expression for Combine the expressions of and solve,

Step 6 of 9

Chapter 8, Problem 49PE is Solved
Step 7 of 9

Textbook: Physics: Principles with Applications
Edition: 6
Author: Douglas C. Giancoli
ISBN: 9780130606204

Physics: Principles with Applications was written by and is associated to the ISBN: 9780130606204. This textbook survival guide was created for the textbook: Physics: Principles with Applications, edition: 6. The full step-by-step solution to problem: 49PE from chapter: 8 was answered by , our top Physics solution expert on 03/03/17, 03:53PM. This full solution covers the following key subjects: helium, nucleus, gold, nuclei, final. This expansive textbook survival guide covers 35 chapters, and 3914 solutions. The answer to “Professional Application Ernest Rutherford (the first New Zealander to be awarded the Nobel Prize in Chemistry) demonstrated that nuclei were very small and dense by scattering helium-4 nuclei (4He) from gold-197 nuclei (197 Au). The energy of the incoming helium nucleus was 8.00×10?13 J , and the masses of the helium and gold nuclei were 6.68×10?27 kg and 3.29×10?25 kg , respectively (note that their mass ratio is 4 to 197). (a) If a helium nucleus scatters to an angle of 120º during an elastic collision with a gold nucleus, calculate the helium nucleus’s final speed and the final velocity (magnitude and direction) of the gold nucleus. (b) What is the final kinetic energy of the helium nucleus?” is broken down into a number of easy to follow steps, and 118 words. Since the solution to 49PE from 8 chapter was answered, more than 332 students have viewed the full step-by-step answer.

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Professional Application Ernest Rutherford (the first New

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